Research Tweet

What is Hybridization?

The concept of hybridization was introduced by scientist pauling. He defined hybridization as redistribution of the energy of orbitals of individual atoms to give new orbitals of equivalent energy. This new orbitals formed are called hybrid orbital.

Features of Hybridization

• Number of hybrid orbitals = number of atomic orbital that get hybridized.

• Hybridized orbitals have equal energy and shape.

• Hybridization takes place with the orbitals of valence shell.

• Hybridisation does not take place in the isolated atom.

Types of Hybridization

sp Hybridization

When one s and one p orbital in same main shell mix together to give new hybrid, orbital called sp hybridised orbital. Forms linear molecules with 180 angle.

Example: all compounds of beryllium such as BeF2, BeCl2

Orbital Hybridization- sp1, sp2, and sp3 Hybridization, Examples 3

sp2 Hybridization

One s + two p orbitals of the same shell = 3 equivalent orbital , new orbitals formed are called sp2 hybrid orbitals.

Angle: 120

Shape: trigonal planar shape

Example: All the compounds of Boron such as BCl3

Orbital Hybridization- sp1, sp2, and sp3 Hybridization, Examples 2

sp3 Hybridization

1 s orbital + 3 p orbitals of same shell = 4 new equivalent orbital. new orbitals formed are called sp3 hybrid orbitals.

Angle: 109°28’ with one another Shape Tetrahedral

Example: ethane (C2H6), methane.

Orbital Hybridization- sp1, sp2, and sp3 Hybridization, Examples 1

sp3d Hybridization

3p orbitals + 1d orbital = 5 orbitals of equal energy. Angle.

Equatorial orbitals: 3 hybrid orbitals lie in the horizontal plane inclined at an angle of 120 Axial orbitals.

The remaining 2 orbitals lie in the vertical plane at 90 degrees plane Shape : trigonal bipyramidal geometry

Example: Phosphorus pentachloride (PCl5)

Orbital Hybridization- sp1, sp2, and sp3 Hybridization, Examples 4

sp3d2 Hybridization

1s + 3p + 2d orbitals = 6 identical hybrid orbitals.

Shape: octahedron.

Angle: 90 degrees to one another.

Example: sulfur hexafluoride SF6

Orbital Hybridization- sp1, sp2, and sp3 Hybridization, Examples 5

sp3d3 Hybridization

1s +3p+3d =5 orbital of same element mix and recast to form hybrid orbitals

Shape: pentagonal bipyramidal geometry

Example: IF7

Orbital Hybridization- sp1, sp2, and sp3 Hybridization, Examples 6

Concept of Hybridization of Carbon Atom

sp Hybridization: When Carbon is bound to two other atoms with the help of two double bonds or one single and one triple bond.

Example: Hybridization of CO2.

sp2 Hybridization: When carbon atom bonding takes place between 1 s-orbital with two p orbitals then the formation of two single bonds and one double bond between three atoms takes place.

Example: Hybridization of graphite

sp3 Hybridization: When the carbon atom is bonded to four other atoms.

Example: Hybridization of CH4 (Methane)

Orbital Hybridization- sp1, sp2, and sp3 Hybridization, Examples 7
How to Determine Hybridization?
Hybridization of Nitrogen in Ammonia, NH3

Step 1: Write the Lewis structure

The valency of nitrogen is 3.

Thus, it forms 3 bonds with three hydrogen atoms.

There is also a lone pair on nitrogen.

Orbital Hybridization- sp1, sp2, and sp3 Hybridization, Examples 8

Step 2: Calculate number of sigma (σ) bonds

Nitrogen in ammonia is bonded to 3 hydrogen atoms.

Number of sigma bonds are = 3.

Step 3: Calculate number of lone pairs

Number of lone pairs on nitrogen atom = (v – b – c) / 2

= (5 – 3 – 0) / 2 = 1 lone pair

There are 5 balance electrons in nitrogen atom before bond formation.

Step 4: Calculate steric number of nitrogen atom

Steric number = number of σ bonds + number of lone pairs

Thus, 3 + 1 = 4

Step 5: Give hybridization and shape of molecule

Nitrogen in ammonia is sp3 hybridization.

The shape is pyramidal.

Orbital Hybridization- sp1, sp2, and sp3 Hybridization, Examples 9
Hybridization and Shape of XeF4

Step 1: Calculate the number of sigma (σ) bonds

Number of sigma bonds formed by xenon = four as it is bonded to only 4 fluorine atoms.

The valency of fluorine = one.

Step 2: Calculate number of lone pairs

The number of lone pairs on xenon atom = (v – b – c) / 2 = (8 – 4 – 0) / 2

Orbital Hybridization- sp1, sp2, and sp3 Hybridization, Examples 1.4

Step 3: Calculate the steric number of central atom

Steric number = number of σ-bonds + number of lone pairs = 4 + 2 = 6

Step 4: Allocate hybridization and shape of molecule

The hybridization is sp3d2.

Thus, Structure is based on octahedral geometry

Hence, Shape is square planar.

Orbital Hybridization- sp1, sp2, and sp3 Hybridization, Examples 1.3
Hybridization and Shape of SO2

Step 1: Write the Lewis structure

Sulfur’s valency may be 2 or 4 or 6.

Oxygen’s valency = one.

Each oxygen makes two bonds with sulfur atom.

One is sigma bond and the second one is pi bond.

The total number of bonds formed by sulfur with two oxygen atoms = four.

Orbital Hybridization- sp1, sp2, and sp3 Hybridization, Examples 1.2

Step 2: Calculate number of sigma (σ) bonds

The number of sigma bonds formed by sulfur atom = two

As it is bonded to two oxygen atoms.

Step3: Calculate number of lone pairs

The number of lone pairs on sulfur atom is = (v – b – c) / 2 = (6 – 4 – 0) / 2; Thus 1.

Number of valence electrons in sulfur is 6.

Total number of bonds including sigma and pi bonds is = 4.

Step 4: Calculate steric number of central atom

Steric number = no. of σ-bonds + no. of lone pairs = 2 + 1 Thus 3

Step 5: Give the hybridization and shape of molecule

The hybridization is sp2.

Structure is built on trigonal planar geometry with one lone pair.

Shape is = angular.

Orbital Hybridization- sp1, sp2, and sp3 Hybridization, Examples 1.1
Steric Number

If steric number is 4, then it is sp3

If steric number is 3 then it is sp2

If steric number is 2 then it is sp

C1 – SN = 3 (three atoms connected), thus sp2

C2 – SN = 3 (three atoms connected), thus sp2

O4 – SN = 3 (1 atom + 2 lone pairs), thus sp2

O5 – SN = 4 (2 atoms + 2 lone pairs), thus sp3

C6 – SN = 4 (4 atoms), thus it is sp3

C7 – SN = 4 (4 atoms), thus it is sp3

N8 – SN = 4 (3 atoms + 1 lone pair), thus it is sp3

C9 – SN = 2 (2 atoms), thus it is sp

C10 – SN = 2 (2 atoms) thus it is sp

​Hybridization Citations

Share

Share on facebook
Share on twitter
Share on linkedin
Share on whatsapp
Share on email
Share on telegram
Share on google
Share on pinterest
Share on vk
Share on odnoklassniki
Share on tumblr
Share on pocket
Similar Post:

0 0 votes
Article Rating
Subscribe
Notify of
guest
0 Comments
Inline Feedbacks
View all comments
0
Would love your thoughts, please comment.x
()
x